4f4. Finding Nd2+ configuration: Nd (Z = 60) ground state: [{Xe}]\, 4f^4\, 6s^2 Forming Nd2+: Remove 2 electrons from the outermost shell (6s²): {Nd}^{2+}: [{Xe}]\, 4f^4 The 6s electrons are removed first (outermost), leaving the 4f electrons intact. Verification: - Nd: [Xe]\, 4f4\, 6s2 → remove 6s2 → [Xe]\, 4f4 Answer: Option (2) — 4f4

JEE Main · 2023 · Shift-IeasyDNF-039

Nd2+ = ____

d & f-Block Elements · Class 12 · JEE Main Previous Year Question

Question

$\text{Nd}^{2+}$ = ____

Options

a
$4f^2\, 6s^2$
b
$4f^4$
✓
c
$4f^3$
d
$4f^4\, 6s^2$

Correct Answerb

$4f^4$

Detailed Solution

Finding $\text{Nd}^{2+}$ configuration:

Nd (Z = 60) ground state: $[\text{Xe}]\, 4f^4\, 6s^2$

Forming $\text{Nd}^{2+}$ : Remove 2 electrons from the outermost shell (6s²): $\text{Nd}^{2+}: [\text{Xe}]\, 4f^4$

The 6s electrons are removed first (outermost), leaving the 4f electrons intact.

Verification:

Nd: $[\text{Xe}]\, 4f^4\, 6s^2$ → remove $6s^2$ → $[\text{Xe}]\, 4f^4$

Answer: Option (2) — $4f^4$

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